1. 106 = ×26 56 64 15 625 = ×
2. Multiply by 1000 and divide by 8: 121 110 090 807 060 504 000 8 968 880 726 456 484 032 000
The numbers must be small (or else the product will be smaller than the sum), so suppose three of them are equal to 1, say 1; 1; 1; x; y
Next: Suppose two are equal to 1 and one is equal to 2:
Then 4 2
4 2 2 8
(2 1)(2 1) 9
2 1 1 and 2 1 9
2 1 3 and 2 1 3
[
]
9. M× =A 12 (1)
Suppose the string starts as 68517, then it cannot go any further. So it must start as:
69234 69234 69234 … i.e. cycles of length 5.
Since 2006= + + + + +5 5 5 L 5 1, the 2006th digit is the same as the first, so it is 6. The number then is: 69234 6923469234 …69234692346,
2 2 2 2 2 or 4 –1 2 –1 4
2 –9 2 –9 2 –1 –3 1 –3 –1
2 2 2 2 2 2 1 0 1 2
2 –9 2 –9 2 –1 –3 1 –3 –1
2 2 2 2 2 4 –1 2 –1 4
Every 2 by 2 block contains one –9 and three 2’s, hence sum:
–9 + 6 = –3 < 0.
Since there are four –9’s and 21 2’s, the total sum is 21 2 4( 9)× + − = >6 0.
The example and brief explanation are all that is required.
12.
Let’s first consider the bottom horizontal line (as in the sketch).
There are s – 1 of the “small” triangles OX1X2; OX2X3; … ; OXs – 1Xs.
There are s – 2 of the triangles obtained by glueing two small ones together, i.e. OX1X3; OX2X4; … ; OXs – 2Xs.
Then s – 3 of those where three small ones are glued together, etc.
Finally, there is 1 triangle OX1Xs where all small ones are glued together.
So the bottom horizontal line determines (s – 1) + (s – 2) + … + 2 + 1 = 1 ( 1)
2s s− triangles. Each horizontal line determines this many triangles, so we have a total of 1 ( 1)
2hs s− triangles.
X1 X2 X3 Xs
O
13. The smallest possible value for OAK is 102 (we want it small to maximize the number of copies).
We conclude that when OAK = 103, we get the maximum number of copies giving: 103 103+ + +L 103=9785=PINE, i.e. 95 copies of OAK.
14. Let L1L2 and L3L4 be the cuts made by them. The dotted lines indicate the diameters parallel to these lines. Letters A to I indicate the areas of the regions in which they are written. Note that
B=E + H and D=E + F, so that B≥H and D≥F. Now slice L L1 3+slice L L2 4 has area A + B + D + E + I ≥ A + H + E + F + I = 1
2 area of pizza, since B≥H and D≥F. Andy should either choose slice
1 3 2 4
L L or slice L L , (slice P or R in original pizza). Andy should choose slices L L and L L1 3 2 4 (slices P and R in the original pizza)
OR
Draw two extra lines symmetrically as shown Then: P1 = S, P2 = Q2 and Q1 = R
1 2 3 1 2 3
(P + R) (Q + S) (P + P + P + R) (Q + Q + S) P 0
∴ − = − = >
Andy should choose slices P and R.
OR:
and 10 represents a digit
